fraction powerFractionPower2005-08-24 04:41:292009-06-25 18:35:16Definitionexponentialfractional number exponent% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Let $m$ be an integer and $n$ a positive factor of $m$.\, If $x$ is a positive real number, we may write the identical equation
$$(x^{\frac{m}{n}})^n = x^{\frac{m}{n}\cdot n} = x^m$$
and therefore the definition of \PMlinkname{$n^\mathrm{th}$ root}{NthRoot} gives the \PMlinkescapetext{formula}
\begin{align}
\sqrt[n]{x^m} = x^{\frac{m}{n}}.
\end{align}
Here, the exponent $\frac{m}{n}$ is an integer.\, For enabling the validity of (1) for the cases where $n$ does not divide $m$ we must set the following
\textbf{Definition.}\, Let $\frac{m}{n}$\, be a fractional number, i.e. an integer $m$ not divisible by the integer $n$, which latter we assume to be positive.\, For any positive real number $x$ we define the\, {\em fraction power}\, $x^{\frac{m}{n}}$ as the $n^\mathrm{th}$ \PMlinkescapetext{root}
\begin{align}
x^{\frac{m}{n}} := \sqrt[n]{x^m}.
\end{align}
\textbf{Remarks}
\begin{enumerate}
\item The existence of the \PMlinkescapetext{root} in the right hand side of (2) is proved \PMlinkname{here}{ExistenceOfRoot}.
\item The defining equation (2) is independent on the form of the exponent $\frac{m}{n}$:\, If\, $\frac{k}{l} = \frac{m}{n}$,\, then we have\, $(\sqrt[n]{x^m})^{ln} = [(\sqrt[n]{x^m})^n]^l = x^{lm} = x^{kn} =
[(\sqrt[l]{x^k})^l]^n = (\sqrt[l]{x^k})^{ln}$,\, and because the mapping\,
$y\mapsto y^{ln}$\, is injective in $\mathbb{R}_+$, the positive numbers $\sqrt[l]{x^k}$ and $\sqrt[n]{x^m}$ must be equal.
\item The fraction power function\, $x\mapsto x^{\frac{m}{n}}$\, is a special case of power function.
\item The presumption that $x$ is positive signifies that one can not identify all $n^\mathrm{th}$ \PMlinkname{roots}{NthRoot} $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$.\, For example, $\sqrt[3]{-8}$ equals $-2$ and\,
$\frac{2}{6} = \frac{1}{3}$,\, but one \textbf{must not} \PMlinkescapetext{calculate}
$$(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2}
= \sqrt[6]{64} = 2.$$
The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$.\, Here we have\, $l = 6$\, and the mapping\,
$y\mapsto y^{ln}$\, is not injective in\, $\mathbb{R}_-\cup\mathbb{R}_+$.\,
--- Nevertheless, some people and books may use also for negative $x$ the equality\, $\sqrt[n]{x} = x^{\frac{1}{n}}$\, and more generally\, $\sqrt[n]{x^m} = x^{\frac{m}{n}}$\, where one then insists that\, $\gcd(m,\,n) = 1.$
\item According to the preceding item, for the negative values of $x$ the derivative of \PMlinkname{odd roots}{NthRoot}, e.g. $\sqrt[3]{x}$, ought to be calculated as follows:
$$\frac{d\sqrt[3]{x}}{dx} = \frac{d(-\sqrt[3]{-x})}{dx} =
-\frac{d(-x)^\frac{1}{3}}{dx} = -\frac{1}{3}\cdot(-x)^{-\frac{2}{3}}(-1) =
\frac{1}{3\sqrt[3]{(-x)^2}} = \frac{1}{3\sqrt[3]{x^2}}$$
The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$'s, although the \PMlinkescapetext{odd} root functions are not special cases of the power function.
\end{enumerate}